CH 5-2: “R” & “S” Configurations

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Presentation transcript:

CH 5-2: “R” & “S” Configurations 1. Assign a numerical “priority” (high…1...2…3…4…low) to each atom or group bonded to a chiral carbon. The relative priorities are based on atomic number. This system assumes that the lowest priority atom (lowest atomic number) is facing “back”. C2 ? C4 ? IUPAC name? 3-chloro-4-ethyl hexane (1) (4)

clockwise = “R” enantiomer counter clockwise = “S” enantiomer 2. If two atoms are the same, go to the next set of atoms. C2 (H,H,C) C4 (H,C,C) (3) (2) (S)-3-chloro-4-ethyl hexane (4) (1) 3. Determine if the configuration of priorities1-2-3 is clockwise or counter clockwise: clockwise = “R” enantiomer counter clockwise = “S” enantiomer

Give an IUPAC name for the following compound Give an IUPAC name for the following compound. Note that the lowest priority atom is facing “forward”. (1) (4) We see counter-clockwise with the H forward, but if we were behind the molecule we would see clockwise with the H back. ● (2) (3) (R)- 1-bromo-1-cyclopentyl-3-methyl butane For practice, draw the structure of (S)-1-bromo-1-cyclopentyl-3-methyl butane FYI: If a reaction produces a mixture of both enantiomers, this is called a “Racemic Mixture”.